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1296 Chapter 29 | Introduction to Quantum Physics
Figure 29.9 Photoelectric effect. A graph of the kinetic energy of an ejected electron, , versus the frequency of EM radiation impinging on a certain material. There is a threshold frequency below which no electrons are ejected, because the individual photon interacting with an individual electron has insufficient energy to break it away. Above the threshold energy, increases linearly with , consistent with .
The slope of this line is —the data can be used to determine Planck’s constant experimentally. Einstein gave the first successful explanation of such data by proposing the idea of photons—quanta of EM radiation.
Einstein’s idea that EM radiation is quantized was crucial to the beginnings of quantum mechanics. It is a far more general concept than its explanation of the photoelectric effect might imply. All EM radiation can also be modeled in the form of photons, and the characteristics of EM radiation are entirely consistent with this fact. (As we will see in the next section, many aspects of EM radiation, such as the hazards of ultraviolet (UV) radiation, can be explained only by photon properties.) More famous for modern relativity, Einstein planted an important seed for quantum mechanics in 1905, the same year he published his first paper on special relativity. His explanation of the photoelectric effect was the basis for the Nobel Prize awarded to him in 1921. Although his other contributions to theoretical physics were also noted in that award, special and general relativity were not fully recognized in spite of having been partially verified by experiment by 1921. Although hero-worshipped, this great man never received Nobel recognition for his most famous work—relativity.
Example 29.1 Calculating Photon Energy and the Photoelectric Effect: A Violet Light
(a) What is the energy in joules and electron volts of a photon of 420-nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420-nm violet light, given that the binding energy (or work function) of electrons for calcium metal is 2.71 eV?
Strategy
To solve part (a), note that the energy of a photon is given by . For part (b), once the energy of the photon is calculated, it is a straightforward application of to find the ejected electron’s maximum kinetic energy, since
BE is given.
Solution for (a)
Photon energy is given by
(29.6) Since we are given the wavelength rather than the frequency, we solve the familiar relationship for the frequency,
yielding
Combining these two equations gives the useful relationship

Now substituting known values yields

Converting to eV, the energy of the photon is
(29.7)
(29.8)
(29.9)
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